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Taylor series. I'm looking for the first few coefficients (see below). I got the first two right, then I broke it. :P
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\[\frac{3x}{1+x} = \sum_{n=0}^{\infty} c_nx^n\]
I did this first :\[c_1=\frac{f'(x)}{1!}=3\] \[c_2=\frac{f''(x)}{2!}=-3\]So far so good. Now: \[c_3=\frac{f'''(x)}{3!}\neq\frac{3}{2}\] Am I just making an arithmetic mistake, or am I misapplying the formula?
i think you did not calculate f'''(x) correctly, what did u get f'''(x) as ?
because i am getting it as +3
I had\[\frac{f'''3}{3!} = \frac{9(1+x)^{-4}}{3!}\]
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hmm, actually, \( f'''(x) = 18 (1+x)^{-4}\)
Oopsie. :)
did you find your error? or still need help
Looking.
Ok, yeah, that was it. Thanks!
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welcome ^_^
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